Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AIME I Problems/Problem 1"
Line 7: Line 7:
 
Let <math>R(x)=P(x)+Q(x).</math> Since the <math>x^2</math>-terms of <math>P(x)</math> and <math>Q(x)</math> cancel, we conclude that <math>R(x)</math> is a linear polynomial.
 
Let <math>R(x)=P(x)+Q(x).</math> Since the <math>x^2</math>-terms of <math>P(x)</math> and <math>Q(x)</math> cancel, we conclude that <math>R(x)</math> is a linear polynomial.
  
−
Note that the graph of <math>R(x)</math> passes through <math>(16+16,54+54)=(32,108)</math> and <math>(20+20,53+53)=(40,106),</math>
+
Note that
 +
<cmath>\begin{alignat*}{8}
 +
R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\
 +
R(20) &= P(20)+Q(20) &&= 53+53 &&= 106,
 +
\end{alignat*}</cmath>
 +
so the slope of <math>R(x)</math> is <math>\frac{106-108}{20-16}=-\frac12.</math>
 +
 
 +
It follows that the equation of <math></math>R(x)<math> is </math>R(x)=-\frac12x+c<math></math> for some constant <math>c.</math> We substitute <math>x=20</math> into this equation to get <math>c=116.</math>
 +
 
 +
Therefore, the answer is <math>R(0)=c=\boxed{116}.</math>
 +
 
 +
~MRENTHUSIASM
  
 
==Solution 2 (Quadratic Polynomials)==
 
==Solution 2 (Quadratic Polynomials)==

Revision as of 15:57, 17 February 2022

Problem 1

Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$

Solution 1 (Linear Polynomials)

Let $R(x)=P(x)+Q(x).$ Since the $x^2$-terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.

Note that \begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \end{alignat*} so the slope of $R(x)$ is $\frac{106-108}{20-16}=-\frac12.$

It follows that the equation of $$ (Error compiling LaTeX. Unknown error_msg)R(x)$is$R(x)=-\frac12x+c$$ (Error compiling LaTeX. Unknown error_msg) for some constant $c.$ We substitute $x=20$ into this equation to get $c=116.$

Therefore, the answer is $R(0)=c=\boxed{116}.$

~MRENTHUSIASM

Solution 2 (Quadratic Polynomials)

Let \begin{align*} P(x) &= 2x^2 + ax + b, \\ Q(x) &= -2x^2 + cx + d, \end{align*} for some constants $a,b,c$ and $d.$

We are given that \begin{alignat*}{8} P(16) &= 512 + 16a + b &&= 54, \hspace{20mm}&&(1) \\ Q(16) &= -512 + 16c + d &&= 54, \hspace{20mm}&&(2) \\ P(20) &= 800 + 20a + b &&= 53, \hspace{20mm}&&(3) \\ Q(20) &= -800 + 20c + d &&= 53, \hspace{20mm}&&(4) \end{alignat*} and we wish to find \[P(0)+Q(0)=b+d.\] We need to cancel $a$ and $c.$ Since $\operatorname{lcm}(16,20)=80,$ we subtract $4\cdot[(3)+(4)]$ from $5\cdot[(1)+(2)]$ to get \[b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.\] ~MRENTHUSIASM

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_1&oldid=171115"